I can t see my title again!! I m so sad that I still haven t replied to me after sending this for th

The curve is an ellipse: x 2 8 + y 2 4 = 1, apparently the fixed point p is outside the ellipse.

Let p line l: y=k(x-4)+1=kx + 1-4k).

Let a(x1, y1), b(x2, y2), x1 >0->16k2 (16k2 -8k + 1) -4(1+2k2)(32k2-16k-6)=16k2-8(16k2-8k-3-6k2).

64k 2+64k+24>0, i.e. 8k 2-8k-3<0, i.e. (2k-1) 2<5 2, so - 10 2 < 2k-1 < 10 2

So (2- 10) 4 < k < 2+ 10) 4

x1+x2= 2k(4k-1)/(1+2k^2) ,x1x2=2(16k^2-8k-3)/(1+2k^2)

Since pa pb = aq qb

So (4-x1) (4-x2) = (x0-x1) (x2-x0).

So (4-x1)(x2-x0) = (x0-x1)(4-x2).

So 4x2 - x1x2 -4x0 + x0x1 = 4x0 - x0x2 - 4x1+ x1x2

So 8x0 - x0(x1+x2) = 4(x1+x2) -2x1x2

So x0(8 - 2k(4k-1) (1+2k 2)) = 4(2k(4k-1) (1+2k 2)) 4(16k 2-8k-3) (1+2k 2).

So x0=[4k(4k-1) -2(16k 2-8k-3)] [4(1+2k2)-k(4k-1)].

16k^2+12k+6]/[4k^2+k+4]=-4+(16k+22)/[4k^2+k+4]

y0=k(x0-4)+1=1-8k+(16k^2+22k)/[4k^2+k+4]=5-8k+(18k-16)/[4k^2+k+4]

32k^3+12k^2-9k+4]/[4k^2+k+4]

When k=0, x0=3 2, y0=1

When k=1, x0=2 9, y0=-25 9

k=1 2, x0=16 11, y0=-3 11

When k = 1 4, x0 = 16 9, y0 = 4 9

It seems that I can't find any rules, and I can't find a way to eliminate k, maybe the calculation is wrong.

We haven't learned this kind of topic yet, but seeing that you are so hardworking, I helped you ask the great god! He did, but I don't know if you can read the blue one.

It's not that no one answered. It's not going to do.

The point is, no! How.

I'm a sophomore in high school this year, and I can look at chemistry questions.

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Question 9: Since ae bisects bad, bae=45°, and abe is a right angle, so aeb=180-90-45=45°, therefore, abe is an isosceles right triangle, so ab=be;

and aod=120°, so aob=180-120=60°, and the angles formed by the two diagonals are equal, so abo= ba0=(180-60) 2=60°, so abo is an equilateral triangle, so bo=ab;

With 2, we know that obc= ocb=60 2=30° (aob is their outer angle).

Synthesis 1,2, we can see that be=bo, and obc=30°, so boe=(180-30) 2=75°;

Question 10: Since bfc and bec are both right-angled triangles, and the common sides bc and m are the midpoints, the points e, f, b, and c are concircular (the center of the circle is the point m).

So fm=em=radius=bc 2=4

So the circumference of fem=4+4+5=13

Pure hand-played, welcome to communicate

Kid, this won't be, just take care of falling in love in class.